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My Notes Ask Your Teacher Values of modulus of elasticity (MOE, the ratio of stress, .e., force per unit area, to strain, .e., deformation per unit length, in GPa) and flexural strength (a measure of the ability to resist failure in bending, in MPa) were determined for a sample ㅁfconcrete beams cf a !rtein typer resulting in the following data MOE 97 33.1 33.8 35.3 35.7 36.2 36.3 36.4 37.7 37.7 3B.6 38.6 39.6 41.1 Strength 5.8 7.0 7.2 6.3 B.0 6.6 7.1 7.5 6.6 6.7 7.2 6.4 B.1 g.1 MOE 42.7 42.8 43.6 45.6 45.8 47.0 47.8 49.2 51.6 62.5 69.9 79.6 8D.1 Strength . 8. B. 9.5 7.37.5 9.9 7.6 7.6 11.5 11.2 11.9 10.8 Fitting the simple linear regression model to the n-27 observations on x-modulus of elasticity and y-nexural strength given in the data above resulted in穸ー7.555, sy-0.179 when x = 40 and夕= 9.731' S-一0.253 ror x-5D. (a) Explain why sy Is larger whenx-60 than when x-40. The coser x is bo y, the saller the vale of The farther x is from x, the smaller the value of The farther x is from y, the smaller the value of O The closcr x is to x, the smaller the valuc of s alt late a c nfidence interval with ล confidence level tof 95 for the true a erage strength of all beams whase modulus o e as c y 1s 40. Round your answers o three der mal Flaces. b the strength of a sin (c) Calculate a prediction interval with a prediction level or 95% for MPa gle beam whose modulusof elasticity is 40· [Round your answers to three decimal places.) f a 95% Cl s calculated or truc average strength when modulus of c asticity 1s 60 what will be the simultaneous confidence ove or both his interval and the interval calcuated in part b ? d The simultancous confidenc level for thcse intervals is at lcast Need Help?Reat Talk to Tutor

Answer 1

b)

X Value=   40      
Confidence Level=   95%      
          
          
Sample Size , n=   27      
Degrees of Freedom,df=n-2 =   25      
critical t Value=tα/2 =   2.060   [excel function: =t.inv.2t(α/2,df) ]  

standard error, S(ŷ)=    0.179              
margin of error,E=t*Std error=t* S(ŷ) =   2.0595   *   0.1789   =   0.3684
                  
Confidence Lower Limit=Ŷ +E =    7.555   -   0.3684   =   7.187
Confidence Upper Limit=Ŷ +E =   7.555   +   0.3684   =   7.923

c)

X̅ =    45.11
Σ(x-x̅)² =Sxx   4582.9
Standard Error of the Estimate,Se=   0.87

standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   0.8835              
margin of error,E=t*std error=t*S(ŷ)=    2.0595   *   0.88   =   1.8196
                  
Prediction Interval Lower Limit=Ŷ -E =   7.555   -   1.8196   =   5.735
Prediction Interval Upper Limit=Ŷ +E =   7.555   +   1.8196   =   9.375

d)

the simultaneous confidence interval is at least 100(1-2α) = 100(1-2*0.05) = 90%